Problem 33 - Calculus Early Transcendentals

Question

Let \(s(t)=5 t^{2}-22 t\) be the position function of a particle moving along a coordinate line, where \(s\) is in feet and \(t\) is in seconds. (a) Find the maximum speed of the particle during the time interval \(1 \leq t \leq 3\). (b) When, during the time interval \(1 \leq t \leq 3,\) is the particle farthest from the origin? What is its position at that instant?

Step-by-Step Solution

Answer

(a) Max speed is 12 ft/s at \(t=1\). (b) Farthest from origin at \(t=1\), position \(-17\) feet.

Step 1: Find the Velocity Function

The velocity function is the derivative of the position function. Given \(s(t) = 5t^2 - 22t\), differentiate it with respect to \(t\) to find the velocity, \(v(t)\): \[ v(t) = \frac{d}{dt}(5t^2 - 22t) = 10t - 22.\]

Step 2: Maximize the Speed within the Given Interval

Speed is the absolute value of velocity, \(|v(t)|\). To find the time when speed is maximized on \(1 \leq t \leq 3\), consider critical points and endpoints. Solve \(v(t) = 0\) to find critical points:\[ 10t - 22 = 0 \quad \Rightarrow \quad t = 2.2. \]Evaluate speed at \(t = 1\), \(t = 2.2\), and \(t = 3\):\[ |v(1)| = |10(1) - 22| = 12, \]\[ |v(2.2)| = |10(2.2) - 22| = 0, \]\[ |v(3)| = |10(3) - 22| = 8. \]The maximum speed is 12 ft/s, occurring at \(t=1\).

Step 3: Determine Particle's Farthest Position from Origin

To find when the particle is farthest from the origin, analyze \(|s(t)|\) within the given interval. Calculate \(s(t)\) at \(t=1\), \(t=2.2\) (from Step 2), and \(t=3\):\[ s(1) = 5(1)^2 - 22(1) = -17, \]\[ s(2.2) = 5(2.2)^2 - 22(2.2) = -7.4,\]\[ s(3) = 5(3)^2 - 22(3) = -3. \]The particle is farthest from the origin at \(t=1\), with a position of \(-17\) feet.

Key Concepts

Velocity Function

The velocity function is a fundamental concept in calculus, especially when studying motion along a line. It gives us a detailed picture of how fast an object is moving in a specific direction at any given time.
In mathematical terms, the velocity function is the derivative of the position function. For a given position function, such as \(s(t) = 5t^2 - 22t\), the velocity function \(v(t)\) is found by differentiating \(s(t)\) with respect to \(t\). This provides us with \(v(t) = 10t - 22\). Here, differentiation tells us the rate at which the position changes, which is the essence of velocity.
This function helps assess not only how fast something moves but also in which direction. A positive velocity indicates motion in the positive coordinate direction, while a negative value suggests the opposite. This is crucial for determining changes in movement over time.

Position Function

Understanding the position function is essential since it describes the location of an object at any given time relative to a reference point. In our example, the position function \(s(t) = 5t^2 - 22t\) expresses the distance of a particle along a line as a function of time \(t\).
The position function helps in tracking the trajectory over time: it shows where the particle is at each moment by plugging in specific values of \(t\). The highest or lowest points along this trajectory — where the particle is farthest from the origin — can be determined by examining where the change in position is greatest. These instances align with the maximum or minimum values of the function.
In practical terms, using the position function over a specific interval, like \(1 \leq t \leq 3\), allows us to determine various points along a particle's path and understand how it moves relative to the starting point.

Critical Points

Critical points are pivotal in analyzing functions, as they represent values of \(t\) where the behavior of a function could change significantly. They are derived from setting the first derivative, such as a velocity function, to zero and solving for \(t\).
Critical points help identify where a function like velocity might reach a minimum or maximum — areas of pause or transition. For instance, the velocity function \(v(t) = 10t - 22\) becomes zero at \(t = 2.2\). At this moment, the particle momentarily stops before potentially changing direction.
Evaluating the original position function at these critical points, alongside interval endpoints, allows us to ascertain where the particle is farthest from the origin or where it might reach its highest speed. They are a cornerstone for exploring not just motion, but any function's turning points when analyzing curves or data trends.